HDU 2120--Ice_cream's world I【并查集， 判断环的个数】

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 783    Accepted Submission(s): 458

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

 

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

 

Output

Output the maximum number of ACMers who will be awarded.
One answer one line.

 

Sample Input

8 100 11 21 32 43 40 55 66 73 64 7

 

Sample Output

3

 

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cstdlib>using namespace std;#define maxn 1100int N, sum, M;int per[maxn];void init() {    sum = 0;    for(int i = 0; i < N; ++i)        per[i] = i;}int find(int x){    int r = x;    while( r != per[r])        r = per[r];    per[x] = r;    return r;}void join (int a, int b){    int fa = find(a);    int fb = find(b);    if(fa != fb){        per[fa] = fb;    }    else sum++;}int main (){    while(scanf("%d%d", &N, &M) != EOF){        init();        while(M--){            int a, b;            scanf("%d%d", &a, &b);            join(a, b);        }        printf("%d\n", sum);    }    return 0;}